a)\(\frac{-2n^3+n^2-5n}{2n+1}\)= \(\frac{-n^2\left(2n+1\right)+n\left(2n+1\right)-6n}{2n+1}\)=\(\frac{\left(2n+1\right)\left(2n-1\right)-6n}{2n+1}\)
=\(\left(n-n^2\right)-\frac{6n}{2n+1}\)=\(\left(n-n^2\right)-\frac{3\left(2n+1\right)-3}{2n+1}\)=\(\left(n-n^2\right)-3-\frac{3}{2n+1}\)
Để (-2n3+n2-5n)⋮(2n+1) thì n∈Z
⇒n∈Z thì (2n+1)∈Ư(3)=\(\left\{-1;-3;1;3\right\}\)
Ta có bảng sau:
| 2n+1 | 1 | 3 | -1 | -3 |
| n | 0 | 1 | -1 | -2 |
Vậy n=(0;1;-1;-2) thì (-2n3+n2-5n) chia hết cho (2n+1).
b)\(\frac{3n^3+10n^2-5}{3n+1}\)=\(\frac{n^2\left(3n+1\right)+3n\left(3n+1\right)-\left(3n+1\right)-4}{3n+1}\)
=\(\frac{\left(3n+1\right)\left(n^2+3n-1\right)-4}{3n+1}\)=\(\left(n^2+3n-1\right)-\frac{4}{3n+1}\)
Để (3n3+10n2-5)⋮(3n+1) thì n∈Z
⇒n∈Z thì (3n+1)∈Ư(4)=\(\left\{1;2;4;-1;-2;-4\right\}\)
Ta có bảng sau:
| 3n+1 | 1 | 2 | 4 | -1 | -2 | -4 |
| n | 0 | \(\frac{1}{3}\) | 1 | \(\frac{-2}{3}\) | -1 | \(\frac{-5}{3}\) |
Vì n∈Z nên ta loại (\(\frac{1}{3}\) ;\(\frac{-2}{3}\); \(\frac{-5}{3}\)) .
Vậy n=(0;1;-1) thì (3n3+10n2-5) chia hết cho (3n+1).
chúc bạn học tốt ^_^